Question

Three identical spheres A,B,C , each of radius R,are placed touching one another on a horizontal table.where is the centre of mass of the system located relative to A(say)

Answer 1

Let center of mass of sphere A is (0,0)

so center of mass of sphere B is at (2R,0)

Center of mass of sphere C is at $(R,\sqrt3)$

now for the center of mass of the system is given as

$r = \frac{m_1r_1 + m_2r_2 + m_3r_3}{m_1 + m_2 + m_3}$

$r = \frac{m(0,0) + m(2R,0) + m(R,\sqrt3R)}{m + m + m}$

$r = \frac{(0,0) + (2R, 0) + (R,\sqrt3R)}{3}$

$r = (0,\frac{R}{\sqrt3})$

so above is the position of center of mass

Answer 2

Consider the balls to form a triangular plane.

Let A(0,0)----->CENTER OF MASS OF BALL 1 PLACED AT ORIGIN

     B(2R,0)---->CENTER OF MASS OF BALL 2 PLACED AT A DISTANCE OF 2R FROM BALL A AND ALTITUDE 0....i.e on the x axis.

     C(R, √3)---->CENTRE OF MASS OF BALL C PLACED ON X-Y PLANE

Be the coordinates of the centres of masses of each of the constituent balls of the resultant system.

The system formed by the three balls have coordinates as follows:-

FOR x coordinate,

x = (m₁x₁ + m₂x₂ + m₃x₃) ÷ ( m₁ + m₂ + m₃)

  = {m(0 + 2R + R)} ÷ 3m

  = 3Rm ÷ 3m

  = R

hence x coordinate of centre of mass is R

SIMILARLY SOLVING FOR y coordinates,

"y coordinate" of center of mass is R/√3

HENCE CENTER OF MASS OF THE SYSTEM OF 3 BALLS IN THE FORM OF A PLANAR TRIANGLE ARE ( R, R/√3).